Question

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. An iron rod is removed from a blacksmith's forge at a temperature of \(900^{\circ} \mathrm{C}\). Assume that \(k=0.02\) and the rod cools in a room with a temperature of \(30^{\circ} \mathrm{C}\). When does the temperature of the rod reach \(100^{\circ} \mathrm{C} ?\)

Short answer

Answer: It takes approximately \(81.83\) minutes for the temperature of the iron rod to reach \(100^{\circ}\mathrm{C}\).

Step-by-step solution

Write down the differential equation

According to Newton's Law of Cooling: \(\frac{dT}{dt} = -k (T - T_a)\) where \(T(t)\) is the temperature of the object at time \(t\), \(T_a\) is the ambient temperature, and \(k\) is a positive constant.

Solve the differential equation

To solve the differential equation \(\frac{dT}{dt} = -k (T - T_a)\), we can use separation of variables. First, rewrite the differential equation in the form to make separation easier: \(\frac{1}{T - T_a} \frac{dT}{dt} = -k\). Now, integrate both sides with respect to \(t\): \(\int \frac{1}{T - T_a} \frac{dT}{dt} dt = -k \int dt\). Make the substitution \(u = T - T_a\). Then, \(du = \frac{dT}{dt} dt\). So, the integral becomes: \(\int \frac{1}{u} du = -k \int dt\). Now, find the antiderivatives: \(\ln|u| = -kt + C\), where \(C\) is the constant of integration. Next, solve for \(T\): \(T(t) = T_a + ue^{-kt}\).

Apply the initial conditions and given values

Now we know the temperature function is \(T(t) = T_a + ue^{-kt}\). We are given the initial temperature of the iron rod \(T(0) = 900^{\circ} \mathrm{C}\), the ambient temperature \(T_a = 30^{\circ} \mathrm{C}\), and the constant \(k = 0.02\). We can find \(u\) by solving \(T(0) = T_a + ue^{-k(0)}\): \(900 = 30 + u \Rightarrow u = 870\). Therefore, the temperature function is: \(T(t) = 30 + 870e^{-0.02t}\).

Find the time when the temperature reaches \(100^{\circ} \mathrm{C}\)

We need to find \(t\) when \(T(t) = 100^{\circ}\mathrm{C}\). So \(100 = 30 + 870e^{-0.02t}\). To find \(t\), solve the equation for \(t\): \(70 = 870e^{-0.02t}\) \(\Rightarrow e^{-0.02t} = \frac{70}{870}\) Take the natural logarithm of both sides: \(-0.02t = \ln\left(\frac{70}{870}\right)\). Now, solve for \(t\): \(t = \frac{\ln\left(\frac{70}{870}\right)}{-0.02} \approx 81.83\). So, the temperature of the rod reaches \(100^{\circ}\mathrm{C}\) after approximately \(81.83\) minutes.

Key concepts

Differential Equation

Newton's Law of Cooling involves a differential equation to model how an object's temperature changes over time. This law is expressed as \( \frac{dT}{dt} = -k (T - T_a) \), where:

• \( \frac{dT}{dt} \) is the rate of change of temperature over time.
• \( k \) is a positive constant that depends on the characteristics of the object and the environment.
• \( T_a \) is the ambient temperature or the temperature of the surrounding environment.

This equation tells us that the rate of cooling is proportional to the difference between the object's temperature \( T \) and the ambient temperature \( T_a \). Understanding this relationship is crucial as it helps us predict how quickly an object will approach the environmental temperature.
In our exercise, the differential equation is set up to find when a hot iron rod, initially at \( 900^{\circ} \mathrm{C} \), cools down in a room at \( 30^{\circ} \mathrm{C} \).

Temperature

Temperature is a fundamental aspect of Newton's Law of Cooling, which concerns how an object's temperature approaches the ambient temperature over time. In the given example:

• The initial temperature \( T_0 \) of the iron rod is \( 900^{\circ} \mathrm{C} \).
• The ambient temperature \( T_a \) is \( 30^{\circ} \mathrm{C} \).
• We are interested in finding the time when the rod reaches \( 100^{\circ} \mathrm{C} \).

Temperature essentially drives the entire cooling process, acting as the primary factor that the entire model revolves around. It's the difference between the object's initial temperature and the ambient temperature that dictates how rapidly the object cools. Using the temperature function derived, \( T(t) = 30 + 870e^{-0.02t} \), allows us to calculate the exact time needed to reach specific temperatures.

Separation of Variables

Separation of variables is a powerful method used to solve differential equations. In our exercise, it's employed to solve the equation for Newton's Law of Cooling, \( \frac{dT}{dt} = -k (T - T_a) \). Here's a concise look at the process:

• Rewrite the equation to separate variables: \( \frac{1}{T - T_a} \frac{dT}{dt} = -k \).
• Integrate both sides with respect to \( t \): \( \int \frac{1}{T - T_a} dT = -k \int dt \).
• Substitution and integration: Use \( u = T - T_a \) to simplify and find the antiderivatives resulting in \( \ln|u| = -kt + C \).

This method provides a solution that relates the temperature \( T(t) \) and time \( t \) explicitly. By separating variables and integrating, we effectively break down the problem into manageable parts, allowing us to solve for the temperature as a function of time.

Initial Conditions

Initial conditions are crucial for solving a differential equation uniquely. They provide the necessary information to determine the constants that arise during integration. In our exercise:

• The initial temperature \( T(0) \) is \( 900^{\circ} \mathrm{C} \).
• This informs us that at \( t = 0 \), \( T(t) = 900 \).

Applying these conditions helps us find the constant \( u \) in the temperature function. Substitute \( T(0) = 30 + u \cdot e^{-0.02 \times 0} = 30 + u \) into the equation to solve for \( u = 870 \).
Thus, the temperature function becomes \( T(t) = 30 + 870e^{-0.02t} \). Initial conditions are therefore indispensable, as they lay the foundation for the specific solution tailored to the problem at hand.